Properties of $ f(x)=|x|^{1/(x-1)} $ (MCQ Contest) 
Let $f$ be a function defined by: 
  $$ f(x)=|x|^{\dfrac{1}{x-1}} $$
  Then : 
Choose the correct option. more than one may be correct 
  
  
*
  
*the domain of $f$ is  $D_{f}=\mathbb{R}\backslash \{1\}$  
  
*$f$ can be extended by continuity to $1$ with $f(1)=\dfrac{1}{e}$
  
*$f$ can be extended by continuity to $1$ and extended function is differentiable at $1$ And the derivative is $-\dfrac{e}{2}$
  
*$f$ is differentiable on its domain, and $f'(x)=\dfrac{1}{(x-1)^{2}}|x|^{\dfrac{1}{x-1}-1}$
  

My thoughts


*

*False because 
$D_{f}=\{x\in \mathbb{R}\mid \ f(x)\in \mathbb{R} \}$ then : 
$x\in D_{f} \iff f(x)\in\mathbb{R} \iff x-1\neq 0  \text{ and } x\neq 0 \iff x\neq 1 \text{ and } x\neq 0  \iff D_{f}=\mathbb{R}\backslash \{0,1\}$

*False because
when $x$ goes to $1$ then $|x|=x$
$ \lim_{x\to 1}f(x)= \lim_{x\to 1}|x|^{\dfrac{1}{x-1}}=\lim_{x\to 1}e^{\dfrac{1}{x-1} \ln(x)}=e^{\lim\limits_{x\to 1}\dfrac{1}{x-1} \ln(x)}=e^{1}$ since $x\mapsto e^x$ is continous and  $\lim\limits_{x\to 1}\dfrac{1}{x-1} \ln(x)=1$ then $\lim_{x\to 1}f(x)=e\neq \dfrac{1}{e}$

*true because we looking to calculate $f'(1)$
when $x$ goes to $1$, $|x|=x$ then
$ f(x)=x^{\dfrac{1}{x-1}}=e^{\dfrac{1}{x-1}\ln(x)}$ then 
$ f'(x)=\left(e^{\dfrac{1}{x-1} \ln(x)}\right)'=
\left(\dfrac{ \dfrac{1}{x}(x-1)-\ln(x)}{(x-1)^{2}}\right)e^{\dfrac{1}{x-1} \ln(x) }
=\left(\dfrac{ \dfrac{x-1-x\ln(x)}{x} }{(x-1)^{2}}\right)e^{\dfrac{1}{x-1} \ln(x) }=\left( \dfrac{x-1-x\ln(x) }{x(x-1)^{2}}\right)e^{\dfrac{1}{x-1} \ln(x) }  $


then $\lim_{x\to 1 } f'(1)$
by applying hopital's rule twice we get 
$$\lim _{x\to \:1}\left(\frac{x\left(1-\ln \left(x\right)\right)-1}{x\left(x-1\right)^2}\right)=-\frac{1}{2}=\lim _{x\to \:1}\left(\frac{-\ln \left(x\right)}{\left(x-1\right)\left(3x-1\right)}\right)=\lim _{x\to \:1}\left(\frac{-\frac{1}{x}}{6x-4}\right)=-\frac{1}{2}$$ on the other hand $f(1)=e$ finaly $$f'(1)=-\dfrac{e}{2}$$
 4. False
 since $ f'(x)=\left( \dfrac{x-1-x\ln(x) }{x(x-1)^{2}}\right)e^{\dfrac{1}{x-1} \ln(x) }  \neq \dfrac{1}{(x-1)^{2}}|x|^{\dfrac{1}{x-1}-1}$
From Wolframe i got :



*

*Is my proof correct?
 A: For the domain, you have to recognize that if $x = 0$, then you would have $|0|^{1/(0-1)} = 0^{-1} = 1/0$, which is clearly not well-defined.
If we take the limit as $x \to 1$, in this neighborhood, $|x| = x$, hence we find $$\lim_{x \to 1} f(x) = \exp \left( \lim_{x \to 1} \frac{\log x}{x-1} \right) = \exp\left( \lim_{x \to 1} \frac{\log x - \log 1}{x - 1} \right)$$ from which we conclude that the inner limit is simply the derivative of $g(x) = \log x$ at $x = 1$, thus the value of $f(x)$ that would make $f$ continuous at $x = 1$ is $e^1 = e$.
For (3), observe as in (2) that in the neighborhood of $x = 1$, $|x| = x$ and it suffices to consider $$h(x) = x^{1/(x-1)},$$ and logarithmic differentiation gives $$h'(x) = h(x) \frac{d}{dx}\left[\frac{\log x}{x-1}\right] = x^{1/(x-1)} \left( \frac{x(1-\log x) - 1}{x(x-1)^2} \right).$$  If we "fill in" the point of discontinuity at $x = 1$ by defining $h(1) = e$, then we must consider the limit of the other factor.  We can evaluate this by a number of methods, but L'Hoptial's rule makes quick work of it.  The result is $-1/2$ for the factor in parentheses, hence $h$ has a derivative of $-e/2$ at $x = 1$.
So for at least the first three parts, the answers should be:  (1) False, (2) False, (3) True.
